The vapour pressure of dilute aqueous solution of glucose (C6div2O6) is 750 mm of mercury at 373K. Calculate 1) Molality 2) Mole fraction of the solute

To tackle the problem of calculating the molality and mole fraction of glucose in a dilute aqueous solution, we need to understand a few key concepts related to vapor pressure and colligative properties. Let's break this down step by step.

Understanding Vapor Pressure and Raoult's Law

The vapor pressure of a solution is influenced by the presence of solute particles. According to Raoult's Law, the vapor pressure of a solvent in a solution is directly proportional to the mole fraction of the solvent. When a non-volatile solute like glucose is added to water, the vapor pressure decreases compared to that of pure water.

Given Data

• Vapor pressure of the solution (P_solution) = 750 mmHg
• Vapor pressure of pure water at 373 K (P°_water) = 760 mmHg

Step 1: Calculate the Mole Fraction of the Solvent

Using Raoult's Law, we can express the relationship between the vapor pressures:

P_solution = X_solvent * P°_solvent

Rearranging this gives us:

X_solvent = P_solution / P°_solvent

Substituting the values:

X_solvent = 750 mmHg / 760 mmHg = 0.9868

Step 2: Calculate the Mole Fraction of the Solute

The mole fraction of the solute (X_solute) can be found using the relationship:

X_solute = 1 - X_solvent

Thus:

X_solute = 1 - 0.9868 = 0.0132

Step 3: Calculate Molality

Molality (m) is defined as the number of moles of solute per kilogram of solvent. To find molality, we first need to determine the number of moles of glucose in the solution.

Let’s assume we have 1 kg of water (the solvent). The mole fraction of glucose (X_solute) is 0.0132, which means:

X_solute = moles of solute / (moles of solute + moles of solvent)

For 1 kg of water, the number of moles of water (moles of solvent) is:

moles of solvent = 1000 g / 18 g/mol = 55.56 moles

Now we can set up the equation:

0.0132 = moles of solute / (moles of solute + 55.56)

Let’s denote the moles of solute as n:

0.0132 = n / (n + 55.56)

Cross-multiplying gives:

0.0132(n + 55.56) = n

Expanding this results in:

0.0132n + 0.733 = n

Rearranging yields:

n - 0.0132n = 0.733

0.9868n = 0.733

n = 0.733 / 0.9868 ≈ 0.742 moles

Now, we can calculate molality:

Molality (m) = moles of solute / kg of solvent = 0.742 moles / 1 kg = 0.742 m

Final Results

To summarize:

• Molality of the solution: 0.742 m
• Mole fraction of the solute (glucose): 0.0132

This process illustrates how we can use vapor pressure data to derive important properties of solutions, which is fundamental in physical chemistry. If you have any further questions or need clarification on any of these steps, feel free to ask!